Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Sequence basics in an arbitrary ordered field: limits are unique, limits preserve non-strict inequalities, convergent sequences are Cauchy, Cauchy sequences are bounded, and a Cauchy sequence with a convergent subsequence converges

Statement

Let FF be an ordered field (Ordered field) and let (xk)(x_k), (yk)(y_k) be sequences in FF, with convergence in FF, Cauchyness in FF, boundedness and subsequences as in Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field. Then:

  1. Limits are unique. If xkLx_k \to L and xkLx_k \to L' in FF, then L=LL = L'. A convergent sequence therefore has exactly one limit in FF and the notation limkxk\lim_k x_k denotes it unambiguously. This is the licence under which the remaining clauses are written as equations between limits, and it is not new here: Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field already establishes it, in an arbitrary ordered field and with no completeness or Archimedean hypothesis. It is restated as clause 1 so that this lemma is self-contained as the citation target of the whole abstract chain on this page.

  2. Limits preserve non-strict inequalities. If (xk)(x_k) and (yk)(y_k) both converge in FF and xkykx_k \le y_k for every kk, then

    limkxk    limkyk.\lim_k x_k \;\le\; \lim_k y_k .

  3. Convergent implies Cauchy. If (xk)(x_k) converges in FF, it is Cauchy in FF.

  4. Cauchy implies bounded. If (xk)(x_k) is Cauchy in FF, it is bounded.

  5. A Cauchy sequence with a convergent subsequence converges. If (xk)(x_k) is Cauchy in FF and some subsequence (xnj)(x_{n_j}) converges in FF, then (xk)(x_k) converges in FF as well, and

    limkxk  =  limjxnj.\lim_k x_k \;=\; \lim_j x_{n_j} .

    Both sides are asserted to exist: the right-hand side by hypothesis, the left-hand side as part of the conclusion.

Why this is a separate item. Each of the five is proved in this library for sequences of reals, and none of those proofs may be cited here. Conventions for sequences: indexing, eventually, lim\lim, and rational ε\varepsilon is explicit about it: a theorem about sequences of reals is a theorem about R\mathbb{R}, and the fact that its argument would transfer to an arbitrary ordered field is a statement about the argument, not a licence to cite the result. The five are collected here, proved from the ordered field axioms alone, so that the completeness equivalences of this page have one place to cite instead of five inline reconstructions.

Facts & Assumptions

Given: An ordered field FF and sequences (xk)(x_k), (yk)(y_k) in FF. Each of the five claims is proved under its own stated hypotheses; nothing is assumed of (xk)(x_k) or (yk)(y_k) outside the claim being proved.

[L1]

Sequences in an ordered field: (xk)(x_k) converges to LL in FF when for every ε>0\varepsilon > 0 in FF there is NNN \in \mathbb{N} with xkL<ε|x_k - L| < \varepsilon for all kNk \ge N; (xk)(x_k) is Cauchy in FF when for every ε>0\varepsilon > 0 in FF there is NNN \in \mathbb{N} with xkxl<ε|x_k - x_l| < \varepsilon for all k,lNk, l \ge N; (xk)(x_k) is bounded when there is MFM \in F with xkM|x_k| \le M for every kk; and a subsequence of (xk)(x_k) is a sequence (xnj)jN(x_{n_j})_{j \in \mathbb{N}} for a strictly increasing n:NNn : \mathbb{N} \to \mathbb{N} (Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field).

[L2]

Triangle inequality: u+vu+v|u + v| \le |u| + |v| for u,vFu, v \in F (The triangle inequality).

[L3]

Absolute value: u0|u| \ge 0; u=0|u| = 0 if and only if u=0u = 0; u=u|-u| = |u|; and uuu \le |u| (Basic properties of the absolute value).

[L4]

Order in FF: exactly one of u<vu < v, u=vu = v, v<uv < u holds, so the order is total, and both << and \le are transitive; adding a constant preserves the strict order and two strict inequalities may be added (Order is preserved by adding a constant and by adding inequalities); the nonstrict forms of those two, used below, are the strict forms together with the equality cases, which trichotomy settles (Ordered field).

[L5]

Halving: 0<1F0 < 1_F (The multiplicative identity is positive), so 21F=1F+1F>02 \cdot 1_F = 1_F + 1_F > 0 (Canonical naturals are positive and strictly increasing) and 21F2 \cdot 1_F is nonzero, hence invertible with (21F)1>0(2 \cdot 1_F)^{-1} > 0 (Inverses of positives are positive, and reciprocation reverses order). Writing ε/2\varepsilon/2 for ε(21F)1\varepsilon \cdot (2 \cdot 1_F)^{-1}, an ε>0\varepsilon > 0 gives ε/2>0\varepsilon/2 > 0 and ε/2+ε/2=ε\varepsilon/2 + \varepsilon/2 = \varepsilon (Ordered field).

[L6]

Induction principle on N\mathbb{N} (The principle of mathematical induction).

[L7]

Growth of an index map: a strictly increasing n:NNn : \mathbb{N} \to \mathbb{N} satisfies njjn_j \ge j for every jj (A strictly increasing index map satisfies nkkn_k \ge k).

[L8]

The order on N\mathbb{N} is total and transitive, so of any two indices one is \ge the other, and every index kk satisfies kNk \le N or kNk \ge N (\le is a linear order on N\mathbb{N}).

Proof

technique · direct
1.1

If dFd \in F satisfies d<εd < \varepsilon for every ε>0\varepsilon > 0 in FF, then d0d \le 0: were d>0d > 0, the instance ε=d\varepsilon = d would give d<dd < d, which trichotomy forbids, so d>0d > 0 fails and totality leaves d0d \le 0.

L4algebra
1.2

For every ε>0\varepsilon > 0 in FF one has ε/2>0\varepsilon/2 > 0 and ε/2+ε/2=ε\varepsilon/2 + \varepsilon/2 = \varepsilon.

L5
1.3

Claim 1. Assume xkLx_k \to L and xkLx_k \to L', and let ε>0\varepsilon > 0 in FF be arbitrary; choose N1N_1 with xkL<ε/2|x_k - L| < \varepsilon/2 for kN1k \ge N_1, choose N2N_2 with xkL<ε/2|x_k - L'| < \varepsilon/2 for kN2k \ge N_2, and let NN be whichever of N1,N2N_1, N_2 is the larger.

L1L8choose
1.4

Claim 2. Assume xkLx_k \to L, ykMy_k \to M and xkykx_k \le y_k for every kk, and let ε>0\varepsilon > 0 in FF be arbitrary; choose N1N_1 with xkL<ε/2|x_k - L| < \varepsilon/2 for kN1k \ge N_1, choose N2N_2 with ykM<ε/2|y_k - M| < \varepsilon/2 for kN2k \ge N_2, and let NN be the larger of the two.

L1L8choose
1.5

Claim 3. Assume xkLx_k \to L and let ε>0\varepsilon > 0 in FF be arbitrary; choose NN with xkL<ε/2|x_k - L| < \varepsilon/2 for all kNk \ge N.

L1choose
1.6

Claim 4. For every nNn \in \mathbb{N} there is BFB \in F with xjB|x_j| \le B for all jnj \le n, by induction on nn: for n=0n = 0 take B=x0B = |x_0|; and given such a BB for nn, totality of the order on FF gives either xn+1B|x_{n+1}| \le B, in which case the same BB serves for n+1n + 1, or B<xn+1B < |x_{n+1}|, in which case xn+1|x_{n+1}| serves for n+1n+1 by transitivity.

L1L4L6
1.7

Claim 4, continued. Assume (xk)(x_k) is Cauchy; since 1F>01_F > 0, choose NN with xkxl<1F|x_k - x_l| < 1_F for all k,lNk, l \ge N, so that for kNk \ge N one has xk=(xkxN)+xNxkxN+xN<1F+xN|x_k| = |(x_k - x_N) + x_N| \le |x_k - x_N| + |x_N| < 1_F + |x_N|.

L1L2L4L5choose
1.8

Claim 5. Assume (xk)(x_k) is Cauchy and xnjLx_{n_j} \to L along a strictly increasing nn, and let ε>0\varepsilon > 0 in FF be arbitrary; choose N1N_1 with xkxl<ε/2|x_k - x_l| < \varepsilon/2 for k,lN1k, l \ge N_1, choose N2N_2 with xnjL<ε/2|x_{n_j} - L| < \varepsilon/2 for jN2j \ge N_2, and let NN be the larger of the two, so that nNNN1n_N \ge N \ge N_1 and NN2N \ge N_2.

L1L7L8choose
2.1

For every kNk \ge N in the situation of step 1.3: LL=(Lxk)+(xkL)Lxk+xkL=xkL+xkL<ε/2+ε/2=ε|L - L'| = |(L - x_k) + (x_k - L')| \le |L - x_k| + |x_k - L'| = |x_k - L| + |x_k - L'| < \varepsilon/2 + \varepsilon/2 = \varepsilon.

step 1.2step 1.3L2L3L4
2.2

For every kNk \ge N in the situation of step 1.4: LM=(Lxk)+(xkyk)+(ykM)L - M = (L - x_k) + (x_k - y_k) + (y_k - M), where LxkLxk<ε/2L - x_k \le |L - x_k| < \varepsilon/2 and ykMykM<ε/2y_k - M \le |y_k - M| < \varepsilon/2 and xkyk0x_k - y_k \le 0; adding, LM<εL - M < \varepsilon.

step 1.2step 1.4L3L4
2.3

For all k,lNk, l \ge N in the situation of step 1.5: xkxl=(xkL)+(Lxl)xkL+xlL<ε/2+ε/2=ε|x_k - x_l| = |(x_k - L) + (L - x_l)| \le |x_k - L| + |x_l - L| < \varepsilon/2 + \varepsilon/2 = \varepsilon.

step 1.2step 1.5L2L3L4
2.4

In the situation of steps 1.6 and 1.7, let BB be a bound for xj|x_j| over jNj \le N and set M:=B+1F+xNM := B + 1_F + |x_N|; then Bx00B \ge |x_0| \ge 0 and 1F+xN>01_F + |x_N| > 0, so MBM \ge B and M1F+xNM \ge 1_F + |x_N|, whence xkBM|x_k| \le B \le M for kNk \le N and xk<1F+xNM|x_k| < 1_F + |x_N| \le M for kNk \ge N; as every index satisfies kNk \le N or kNk \ge N, (xk)(x_k) is bounded.

step 1.6step 1.7L1L3L4L8
2.5

For every kNk \ge N in the situation of step 1.8: xkL=(xkxnN)+(xnNL)xkxnN+xnNL<ε/2+ε/2=ε|x_k - L| = |(x_k - x_{n_N}) + (x_{n_N} - L)| \le |x_k - x_{n_N}| + |x_{n_N} - L| < \varepsilon/2 + \varepsilon/2 = \varepsilon, the first summand being covered because kNN1k \ge N \ge N_1 and nNN1n_N \ge N_1.

step 1.2step 1.8L2L4
3.1

By step 2.1 the element LL|L - L'| is below every ε>0\varepsilon > 0, so LL0|L - L'| \le 0; with LL0|L - L'| \ge 0 this forces LL=0|L - L'| = 0 and hence L=LL = L', which is claim 1.

step 1.1step 2.1L3L4
3.2

By step 2.2 the element LML - M is below every ε>0\varepsilon > 0, so LM0L - M \le 0, that is LML \le M, which is claim 2.

step 1.1step 2.2L4
3.3

Step 2.3 produced, for an arbitrary ε>0\varepsilon > 0, an NN beyond which all pairs are within ε\varepsilon, so (xk)(x_k) is Cauchy in FF, which is claim 3.

step 2.3L1
4.1

Step 2.5 produced, for an arbitrary ε>0\varepsilon > 0, an NN beyond which xkL<ε|x_k - L| < \varepsilon, so (xk)(x_k) converges in FF with xkLx_k \to L; since also xnjLx_{n_j} \to L, step 3.1 identifies both limits as LL and gives limkxk=limjxnj\lim_k x_k = \lim_j x_{n_j}, which is claim 5.

step 2.5step 3.1L1
5.1

Claims 1, 2, 3, 4 and 5 are steps 3.1, 3.2, 3.3, 2.4 and 4.1 respectively, so all five hold.

step 2.4step 3.1step 3.2step 3.3step 4.1

Remarks

  • Nothing above uses the Archimedean property, and nothing above uses completeness. The five claims hold in every ordered field, including R(t)\mathbb{R}(t) and R((t1))\mathbb{R}((t^{-1})). That is what makes them safe to use on both sides of every implication proved on this page.

  • Claim 2 is genuinely non-strict. From xk<ykx_k < y_k at every index one gets only LML \le M: the sequences xk=0x_k = 0 and yk=ε/(k+1)y_k = \varepsilon/(k+1) in an Archimedean FF have xk<ykx_k < y_k and equal limits. The real-number version of this warning is recorded at Limits preserve non-strict inequalities.

  • There is deliberately no arithmetic clause here. Nothing above lets one add, multiply or divide two limits in a general ordered field, and no item in this library does: Algebra of limits: sums, scalar multiples, products and quotients is stated for sequences of reals, and by the rule recalled above it may not be cited for a general FF. No proof on this page needs such a clause; every abstract argument here works with the defining ε\varepsilon and NN directly, or with clauses 1 to 5.

  • Claim 4 avoids any appeal to a maximum of a finite set. The library's finite-maximum lemma Every nonempty finite set of reals has a maximum and a minimum is stated for R\mathbb{R}, so it is unavailable here for the same reason the other four real-valued lemmas are; step 1.6 replaces it by an induction that uses nothing but totality of the order of FF.

Depends on

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